2021 AMC Fall 10B Problem 8

The first sentence is a hint: $16{,}383$ is one less than $16{,}384 = 2^{14}$.

1. Write it as a difference of squares. $2^{14} = (2^7)^2 = 128^2$, and $1 = 1^2$. So
$$16{,}383 = 128^2 - 1^2 = (128 - 1)(128 + 1) = 127 \times 129.$$
2. Break down $129$. Its digits add to $12$, so $3$ divides it: $129 = 3 \times 43$. Both are prime.
3. Is $127$ prime? Only primes up to about $\sqrt{127}$ need checking. Since $11 \times 11 = 121$ and $12 \times 12 = 144$, check $2, 3, 5, 7, 11$:
- $127$ is odd, so not $2$.
- Its digits add to $10$, so not $3$.
- It does not end in $0$ or $5$.
- $7 \times 18 = 126$, so not $7$.
- $11 \times 11 = 121$ and $11 \times 12 = 132$, so not $11$.

So $127$ is prime.
4. So $16{,}383 = 3 \times 43 \times 127$, and the largest prime factor is $127$.
5. Its digit sum is $1 + 2 + 7 = 10$.

The answer is $\boxed{\textbf{(C)}\ 10}$.

A number just below a power of $2$ with an even exponent is a square minus $1$, so it splits into two factors right away: $2^{14} - 1 = (2^7 - 1)(2^7 + 1)$. When a problem tells you a fact that seems unrelated, like $16{,}384 = 2^{14}$, it is usually there to point you to exactly this kind of shortcut.

The problem tells us $16384 = 2^{14}$, so $16383 = 2^{14} - 1$, which is a difference of squares:
$$2^{14} - 1 = (2^7 - 1)(2^7 + 1) = 127 \cdot 129.$$
Now $129 = 3 \cdot 43$, and $127$ is prime (it is not divisible by $2, 3, 5, 7$ or $11$, and $11^2 = 121 < 127 < 144 = 12^2$). Thus
$$16383 = 3 \cdot 43 \cdot 127,$$
and the greatest prime divisor is $127$. Its digit sum is $1 + 2 + 7 = 10$.

The answer is $\boxed{\textbf{(C)}\ 10}$.